Question

Sin(7П+x)=cos(9П+2x)

Answer 1

Решение приведено во вложении

Answer 2

We can use the trigonometric identity sin(a+b) = sin(a)cos(b) + cos(a)sin(b) and cos(a+b) = cos(a)cos(b) - sin(a)sin(b) to rewrite the equation:

sin(7π+x) = cos(9π+2x) => sin(7π)cos(x) + cos(7π)sin(x) = cos(9π)cos(2x) - sin(9π)sin(2x)

Since sin(7π) = sin(π) = 0 and cos(7π) = cos(π) = -1, and sin(9π) = sin(π) = 0 and cos(9π) = cos(π) = -1, we can simplify further:

• cos(x) = -sin(2x)
• cos(2x) = -sin(x)

We can use the double angle formula cos(2x) = 1 - 2sin^2(x) to rewrite the second equation:

1 - 2sin^2(x) = -sin(x) => 2sin^2(x) + sin(x) - 1 = 0

We can solve for sin(x) using the quadratic formula:

sin(x) = (-1 ± √5) / 4

Substituting this back into the first equation, we can solve for cos(x):

cos(x) = sin(2x) / (-1) = -cos^2(x) + 1 = -[(-1 ± √5) / 4]^2 + 1

Therefore, the solutions for x that satisfy the equation are:

x = arcsin[(-1 ± √5) / 4] and x = 2π - arcsin[(-1 ± √5) / 4]